Problem: $f(x)=\dfrac{1}{1+{{x}^{2}}}$ Find a power series for $f$. Choose 1 answer: Choose 1 answer: (Choice A) A $1\text{ }+\text{ }x\text{ }+x^2+\ldots +\text{ }{{ x }^{n}}+\ldots$ (Choice B) B $1-x+x^2+\ldots +{{(-1)^n x }^{n}}+\ldots$ (Choice C) C $1+{{x}^{2}}+x^4+\ldots +\text{ }{{ {{x}} }^{2n}}+\ldots$ (Choice D) D $1-{{x}^{2}}+x^4+\ldots +(-1)^n x ^{2n}+\ldots$
Answer: This is a geometric series with first term $a\text{ }=\text{ }1$ and common ratio $r\text{ }=\text{ }{-{x}^{2}}$. Therefore, the series is as follows: $1-{{x}^{2}}+x^4+\ldots +(-1)^n {x} ^{2n}+\ldots $